An arc lamp requires a direct current of 10 A | vedclass

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(B) The resistance of the arc lamp is $R = \frac{V}{I} = \frac{80\ V}{10\ A} = 8\ \Omega$.When connected to an $AC$ supply,the impedance $Z$ of the $R

Vedclass · Accepted Answer

$0.065$

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(B) The resistance of the arc lamp is $R = \frac{V}{I} = \frac{80\ V}{10\ A} = 8\ \Omega$.When connected to an $AC$ supply,the impedance $Z$ of the $RL$ circuit is given by $Z = \sqrt{R^2 + X_L^2}$,where $X_L = 2\pi f L$.The current in the $AC$ circuit is $I = \frac{V_{rms}}{Z} = \frac{V_{rms}}{\sqrt{R^2 + (2\pi f L)^2}}$.Given $I = 10\ A$,$V_{rms} = 220\ V$,$f = 50\ Hz$,and $R = 8\ \Omega$:$10 = \frac{220}{\sqrt{8^2 + (2 \cdot \pi \cdot 50 \cdot L)^2}}$$\sqrt{64 + (100\pi L)^2} = \frac{220}{10} = 22$$64 + (100\pi L)^2 = 22^2 = 484$$(100\pi L)^2 = 484 - 64 = 420$$100\pi L = \sqrt{420} \approx 20.49$$L = \frac{20.49}{100 \cdot 3.14} \approx \frac{20.49}{314} \approx 0.065\ H$.

An arc lamp requires a direct current of $10\ A$ at $80\ V$ to function. If it is connected to a $220\ V$ (rms),$50\ Hz$ $AC$ supply,the series inductor needed for it to work is close to: (in $H$)

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